What is the integral of tan(8x) ?

The integral of tan(8x) is -1/8 ln|cos(8x)|+C

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integral of tan(8x)

\int tan (8 x) d x

- \frac{1}{8} ln ∣ cos (8 x) ∣ + C

Solution steps

\int tan (8 x) d x

Rewrite using trig identities

= \int \frac{sin ( 8 x )}{cos ( 8 x )} d x

Apply u-substitution

: \int - \frac{1}{8 u} d u

= \int - \frac{1}{8 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{8} ln ∣ u ∣

Substitute back

u = cos (8 x)

= - \frac{1}{8} ln ∣ cos (8 x) ∣

Add a constant to the solution

= - \frac{1}{8} ln ∣ cos (8 x) ∣ + C

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